, 2023

Solution:

It is impossible for the whole class to buy tickets to the concert.
If two students $A$ and $B$ are close friends, and $A$ has bought a ticket to the concert while $B$ has not, then $A$ is enticing $B$. We call this pair $(A, B)$ an enticement.
In order for a student to change their mind and buy a ticket, they first be enticed by at least 2 of their 3 close friends. That means they can only entice at most 1 other friend. Therefore, the total number of enticements among the students decreases by 1 whenever a student changes their mind to buy a ticket.

Initially, the maximum number of enticements is 15 (each of the initial 5 students with tickets has 3 friends to entice). Assume, for the sake of contradiction, that the entire class ends up buying tickets. After the first 14 people buy tickets, the number of enticements is at most $15-14=1$. This is not enough to convince the last person to buy a ticket, since they need 2 enticements.
Therefore, it is impossible that the entire class buys tickets.

Solution:

We shall use the term friendship to denote an unordered pair of students who are close friends. Since each of the 20 students is part of exactly 3 friendships, there are exactly 30 friendships in the class. (We could also represent friendships as edges in an undirected graph whose vertices are the 20 students.)
We say that a friendship is used if one of the students in that friendship buys a ticket after the original five buyers, and the other student already has a ticket at that time. Each time a ticket is purchased after the original five purchases, at least two friendships are used. Observe that no friendship gets used twice.

If all 20 students buy tickets, then three friendships are used when the last student buys a ticket. This would imply that the number of used friendships is at least $14 \times 2+3=31$, which is more than the number of friendships. This contradiction proves that it is not possible that the entire class buys tickets.